Problem: $ D = \left[\begin{array}{rr}-1 & 0 \\ 2 & -2 \\ 1 & 3\end{array}\right]$ $ F = \left[\begin{array}{rr}0 & 2 \\ 3 & -2\end{array}\right]$ What is $ D F$ ?
Answer: Because $ D$ has dimensions $(3\times2)$ and $ F$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ D F = \left[\begin{array}{rr}{-1} & {0} \\ {2} & {-2} \\ \color{gray}{1} & \color{gray}{3}\end{array}\right] \left[\begin{array}{rr}{0} & \color{#DF0030}{2} \\ {3} & \color{#DF0030}{-2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-1}\cdot{0}+{0}\cdot{3} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{0}+{0}\cdot{3} & ? \\ {2}\cdot{0}+{-2}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{0}+{0}\cdot{3} & {-1}\cdot\color{#DF0030}{2}+{0}\cdot\color{#DF0030}{-2} \\ {2}\cdot{0}+{-2}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-1}\cdot{0}+{0}\cdot{3} & {-1}\cdot\color{#DF0030}{2}+{0}\cdot\color{#DF0030}{-2} \\ {2}\cdot{0}+{-2}\cdot{3} & {2}\cdot\color{#DF0030}{2}+{-2}\cdot\color{#DF0030}{-2} \\ \color{gray}{1}\cdot{0}+\color{gray}{3}\cdot{3} & \color{gray}{1}\cdot\color{#DF0030}{2}+\color{gray}{3}\cdot\color{#DF0030}{-2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}0 & -2 \\ -6 & 8 \\ 9 & -4\end{array}\right] $